The Flat Earth Society
Flat Earth Discussion Boards => Flat Earth Q&A => Topic started by: sixstringthing on December 06, 2013, 12:38:13 AM
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Hi. I'm a newly confirmed FEer and can't conceive of a RE by logic alone. (William Carpenter complained of having to justify a Flat Earth when it so obviously was.) Logical thought here... How on Earth could gravity hold water in a fixed curvature around a basketball shaped object? It just doesn't make sense and is downright incomprehensible.
My question though is about the "rate" of curvature on the Globe version of Earth. I've read it is 8 inches per mile and then just read a conflicting number from Wilbur Voliva (much higher if I recall).
So please someone, provide a definitive answer, mathematically correct as to the convexity of "planet" Earth. I desperately need this information to verbally abuse knuckleheads who buy into the insane idea that this thing we live on is shaped like a basketball and water is curved.
And Thanks!
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How on Earth could gravity hold water in a fixed curvature around a basketball shaped object?
If I may answer this with a question: which way do you think gravity pulls in RE terms?
My question though is about the "rate" of curvature on the Globe version of Earth. I've read it is 8 inches per mile and then just read a conflicting number from Wilbur Voliva (much higher if I recall).
So please someone, provide a definitive answer, mathematically correct as to the convexity of "planet" Earth.
Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry.
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You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.
11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.
So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?
If the earth was a ball, this is what should realistically happen.
To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.
(http://img43.imageshack.us/img43/231/y8s9.png) (http://imageshack.us/photo/my-images/43/y8s9.png/)
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You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.
11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.
So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?
If the earth was a ball, this is what should realistically happen.
To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.
(http://img43.imageshack.us/img43/231/y8s9.png) (http://imageshack.us/photo/my-images/43/y8s9.png/)
It's very simple. Calculating the curvature of the earth is simply of a parabolic nature. The distance for the first mile is about 8 inches, but increases as the distance increases. It's not linear. Once again sceptimatic seems to be ignorant of the most commonly known facts. Use a ruler and a basketball for example and you can readily see how this works.
There is no confllict if you understand the basic priniciples of math, geometry, trigonometry, physics, etc. But in the words of sceptimatic, they were made up by those mad scientists just to fit a globular earth.
IMHO Simple known facts such as these are where Flat Earth fails.
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I don't think you need to go into all that stuff. It's pretty self explanatory if you do it simply.
If not, show me why I'm wrong, in the usual simple terms.
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Your first question about what direction I think gravity pulls in a RET situation? We are taught that it pulls towards the center of the "globe", or in the general direction of down. I'm not sure why you asked that or why it's relevant to my question?
Your answer as to the "rate" or curvature of the "globe/Earth":
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."
OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.
I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.
With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water. Such experiments have been discussed countless times and stated in different ways. NO "rate of fall" can be calculated anywhere over water.
Conclusion: The Earth is Flat, there can be no other explanation
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You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.
11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.
So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?
If the earth was a ball, this is what should realistically happen.
To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.
(http://img43.imageshack.us/img43/231/y8s9.png) (http://imageshack.us/photo/my-images/43/y8s9.png/)
I don't understand why you need to move back at all. Why, please? I'm just asking a simple, straight forward question that should not involve "moving backwards" from your point of focus.
Can we keep this question simple? How far does Earth drop or "fall" (over water... for simplicity's sake) in one mile?
It appears that the answer is 8 - 8.6 inches/mile and no one has contested that as far as I can determine.
Assuming that 8 - 8.6inches/mile IS the correct answer. No matter which direction over water you measure the fall will be 8 - 8.6 inches/mile.
This obviously reasoning states that "water is not always "flat" and does not seek it's own level"... or else Earth is Flat.
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I was replying to Scintific Method above.
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You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.
11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.
So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?
If the earth was a ball, this is what should realistically happen.
To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.
(http://img43.imageshack.us/img43/231/y8s9.png) (http://imageshack.us/photo/my-images/43/y8s9.png/)
It's very simple. Calculating the curvature of the earth is simply of a parabolic nature. The distance for the first mile is about 8 inches, but increases as the distance increases. It's not linear. Once again sceptimatic seems to be ignorant of the most commonly known facts. Use a ruler and a basketball for example and you can readily see how this works.
There is no confllict if you understand the basic priniciples of math, geometry, trigonometry, physics, etc. But in the words of sceptimatic, they were made up by those mad scientists just to fit a globular earth.
IMHO Simple known facts such as these are where Flat Earth fails.
With all due respect... "Simple known facts such as these" are the reasons why Round Earth becomes and impossibility. (Speaking of simply calculating the rate of drop or fall on a "round" Earth which to this point in the post is agreed to be 8 - 8.6 inches per mile)
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I was replying to Scintific Method above.
Good, because you had me all confused for a second.
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You have to take into account that from your standing point, you are curving away from the curve your are viewing ahead, so it's double each time. So 8 inches for one mile is 4 inches drop from your standing point and 4 inches from your forward view, equals 8 inches.
Move another mile and it doubles to 16 inches, then 32 inches for 3 miles, 64 inches for 4 miles, 128 inches for 5 miles, 256 inches for 6 miles, 512 inches for 7 miles, 1,024 inches for 8 miles, 2,048 inches for 9 miles and 4,096 inches for 10 miles.
So after 10 miles, any view from your point will be 341 feet, approximately, or basically, you have dipped below the curve by 170 and a half feet and the object you look at as dipped the same, so to your view, if say, it was a building that was 342 feet tall...you would see 1 foot of it and that's it.
11 miles goes to 8,192 inches, 12 miles will be 16,384 inches, 13 miles will be 32,768 inches, 14 miles will be 65,536 inches, 15 miles will be 131,072 inches, 16 miles will be 262,144 inches, 17 miles will be 524,288 inches, 18 miles will be 1,048,576 inches, 19 miles will be 2,097,152 inches and 20 miles will be 4,194,304, which would equate to 349,525 FEET or 106,535,METRES, or 66 MILES.
So basically from your point, you will drop down your dip, 33 miles and the thing you are viewing will also drop 33 miles. Do we have any building that can be viewed from 20 miles that are just over 33 miles in height so that we can see the top?
If the earth was a ball, this is what should realistically happen.
To clarify it, let's assume that a person is on a large ball and also there is a model building on it as well. Ok, we know the ball is not the earth, but the principle is the same.
Now let's assume that this ball is 24 feet in circumference.
(http://img43.imageshack.us/img43/231/y8s9.png) (http://imageshack.us/photo/my-images/43/y8s9.png/)
It's very simple. Calculating the curvature of the earth is simply of a parabolic nature. The distance for the first mile is about 8 inches, but increases as the distance increases. It's not linear. Once again sceptimatic seems to be ignorant of the most commonly known facts. Use a ruler and a basketball for example and you can readily see how this works.
There is no confllict if you understand the basic priniciples of math, geometry, trigonometry, physics, etc. But in the words of sceptimatic, they were made up by those mad scientists just to fit a globular earth.
IMHO Simple known facts such as these are where Flat Earth fails.
With all due respect... "Simple known facts such as these" are the reasons why Round Earth becomes and impossibility. (Speaking of simply calculating the rate of drop or fall on a "round" Earth which to this point in the post is agreed to be 8 - 8.6 inches per mile)
Yes, 8 inches per mile but not per mile per mile as in 10 miles being 80 inches.
From your standing point if the curve is 8 inches per mile, then it has to be 4 inches forward and 4 inches back.
It's supposed to be a ball, so either way it has to curve both ways for any person stood on it. I mentioned a person moving back a mile because buildings do not move, but as the person moves back, he is walking down the curve as the building is disappearing down the opposite curve.
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I think we are all agreed then?
The curvature/convexity/"fall" or "drop" of the Globe/Round Earth concept is thus:
For every ONE Mile of travel in any direction (over water, for simplicity) there is a "drop" or "fall" of 8 - 8.6 Inches Per Mile of Travel. As an example, Travel one mile and the drop/fall will be 8 - 8.6 inches. Travel One Hundred miles and the drop/fall will be 800 - 860 Inches, or 66.6 Feet to 71.6 Feet.
Anyone else want to challenge the curvature calculations?
My hope is that we come to a definitive, mathematically proven and undeniable TRUTH as to what the drop or fall of a Round/Globular Earth is.
I think it will then be a simple matter for everyone to accept that Earth cannot possibly be shaped like a basketball or "globe".
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I think we are all agreed then?
The curvature/convexity/"fall" or "drop" of the Globe/Round Earth concept is thus:
For every ONE Mile of travel in any direction (over water, for simplicity) there is a "drop" or "fall" of 8 - 8.6 Inches Per Mile of Travel. As an example, Travel one mile and the drop/fall will be 8 - 8.6 inches. Travel One Hundred miles and the drop/fall will be 800 - 860 Inches, or 66.6 Feet to 71.6 Feet.
Anyone else want to challenge the curvature calculations?
My hope is that we come to a definitive, mathematically proven and undeniable TRUTH as to what the drop or fall of a Round/Globular Earth is.
I think it will then be a simple matter for everyone to accept that Earth cannot possibly be shaped like a basketball or "globe".
I don't agree with the calculations. It fails to take into account what I put in my diagram.
I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile.
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Your first question about what direction I think gravity pulls in a RET situation? We are taught that it pulls towards the center of the "globe", or in the general direction of down. I'm not sure why you asked that or why it's relevant to my question?
It's relevant to the question it was in reply to, as it explains how water can be held in a "fixed curvature", as you put it.
Your answer as to the "rate" or curvature of the "globe/Earth":
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."
OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.
I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.
With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water. Such experiments have been discussed countless times and stated in different ways. NO "rate of fall" can be calculated anywhere over water.
Conclusion: The Earth is Flat, there can be no other explanation
Pay attention. I didn't say "71.6 over 100 miles" I said "71.6 feet per mile". Over 100 miles it would be ~7,160 feet. This can be confirmed by going for a fly (which I have done countless times in my life), and noting that, when at 7,000 feet, you can see about 100 miles. Standing on the ground, you can see for about 3 miles. From about 100 feet above the ground, you can see roughly 12 miles. I've personally verified all of this, and I consider it to be pretty sound proof of curvature. I encourage you to go and verify it for yourself as well, and/or come up with a credible alternative explanation for it.
EDIT: No, I didn't stuff up my calculation, the fact that 71.6 feet divided by 100 is about 8.6 inches is just a coincidence. It threw me off when I first did the maths, so I triple checked it, and it is most definitely right. As an added note, the drop over 6,000 miles is 3,981 miles, or about 3500 feet per mile. If you can tell me why that is, I'll refrain from calling you names. ;)
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Can we keep this question simple? How far does Earth drop or "fall" (over water... for simplicity's sake) in one mile?
In one mile, 8.6 inches.
It appears that the answer is 8 - 8.6 inches/mile and no one has contested that as far as I can determine.
Assuming that 8 - 8.6inches/mile IS the correct answer. No matter which direction over water you measure the fall will be 8 - 8.6 inches/mile.
Please read and understand my explanation in my last post.
This obviously reasoning states that "water is not always "flat" and does not seek it's own level"... or else Earth is Flat.
It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look.
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Your first question about what direction I think gravity pulls in a RET situation? We are taught that it pulls towards the center of the "globe", or in the general direction of down. I'm not sure why you asked that or why it's relevant to my question?
It's relevant to the question it was in reply to, as it explains how water can be held in a "fixed curvature", as you put it.
Your answer as to the "rate" or curvature of the "globe/Earth":
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."
OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.
I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.
With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water. Such experiments have been discussed countless times and stated in different ways. NO "rate of fall" can be calculated anywhere over water.
Conclusion: The Earth is Flat, there can be no other explanation
Pay attention. I didn't say "71.6 over 100 miles" I said "71.6 feet per mile". Over 100 miles it would be ~7,160 feet. This can be confirmed by going for a fly (which I have done countless times in my life), and noting that, when at 7,000 feet, you can see about 100 miles. Standing on the ground, you can see for about 3 miles. From about 100 feet above the ground, you can see roughly 12 miles. I've personally verified all of this, and I consider it to be pretty sound proof of curvature. I encourage you to go and verify it for yourself as well, and/or come up with a credible alternative explanation for it.
EDIT: No, I didn't stuff up my calculation, the fact that 71.6 feet divided by 100 is about 8.6 inches is just a coincidence. It threw me off when I first did the maths, so I triple checked it, and it is most definitely right. As an added note, the drop over 6,000 miles is 3,981 miles, or about 3500 feet per mile. If you can tell me why that is, I'll refrain from calling you names. ;)
"Going for a fly" is only a subjective opinion based on visual assumptions. There can only be ONE mathematically correct number that MUST contiguous for EACH mile of travel on a uniformly shaped object. The original formula can NOT extrapolate out to a different ratio Per Mile on a uniformly round object, right?
Each mile travelled will have the SAME amount of fall (or drop) on a uniformly shaped "ball" or globe.
That amount of drop or fall per mile REMAINS 8 - 8.6 inches per mile for EVERY mile travelled on a "round" Earth, and in fact it doesn't matter which direction you travel in!
The ratio WILL remain constant. It's simple mathematics.
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sceptimatic:
You said: "I don't agree with the calculations. It fails to take into account what I put in my diagram.
I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile."
No, I don't think so. How can the drop/fall possibly double with each mile travelled? No offence, but you need to re-think that one.
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Scintific Method:
You said: "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."
Wow, I'd really like to see water that does not lie flat, and water which is "forming a curve whichever way you look."
Water can not possibly be anything other than flat and seeking it's own level at ALL times.
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sixstringthing, your level of ignorance is staggering, I have only ever seen a couple of other that rival it, and them only on these fora. Perhaps a nice diagram will alleviate it some. Are you at least familiar with gradients?
(http://img23.imageshack.us/img23/4685/as0k.png)
If this were a cross-section of the earth:
- the blue line would represent the drop-per-mile calculated over 1,500 miles (about 1,060 feet per mile);
- the green line would represent the drop-per-mile calculated over 3,000 miles (about 2,040 feet per mile);
- the red line would represent the drop-per-mile calculated over 4,500 miles (about 2,870 feet per mile);
- the yellow line would represent the drop-per-mile calculated over 6,000 miles (as mentioned, about 3,500 feet per mile).
NOW can you see how it is impossible to describe a curve with a single gradient?
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Scintific Method:
You said: "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."
Wow, I'd really like to see water that does not lie flat, and water which is "forming a curve whichever way you look."
Water can not possibly be anything other than flat and seeking it's own level at ALL times.
It's not flat, it's curved, following the curve of the earth, and settling to the lowest points it can (aka the points closest to the centre of the earth, thanks to gravity); finding it's own level so that the surface of any single water mass forms a curve (albeit a very slight curve, not easily perceived with the unaided eye) with a pretty constant radius, the origin point for which is the centre of the earth.
Did you follow that? or do I need to dumb it down even more?
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sceptimatic:
You said: "I don't agree with the calculations. It fails to take into account what I put in my diagram.
I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile."
No, I don't think so. How can the drop/fall possibly double with each mile travelled? No offence, but you need to re-think that one.
If the earth was a globe, which it isn't and you were to drive round it, then yes, every mile you covered you would assume the 8 inch curve. I understand that.
What I'm talking about is by viewing a building and moving back a mile at a time. Each mile moved, creates more curvature BOTH ways as the globe would be uniform, or basically, to save argument of the so called bulge.
Ok, I've re-evaluated. It would only double if two objects were moving away from each other by a mile EACH, not one object being stationary.
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Water can not possibly be anything other than flat and seeking it's own level at ALL times.
put a small drop of water on a flat surface.
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sixstringthing, your level of ignorance is staggering, I have only ever seen a couple of other that rival it, and them only on these fora. Perhaps a nice diagram will alleviate it some. Are you at least familiar with gradients?
(http://img23.imageshack.us/img23/4685/as0k.png)
If this were a cross-section of the earth:
- the blue line would represent the drop-per-mile calculated over 1,500 miles (about 1,060 feet per mile);
- the green line would represent the drop-per-mile calculated over 3,000 miles (about 2,040 feet per mile);
- the red line would represent the drop-per-mile calculated over 4,500 miles (about 2,870 feet per mile);
- the yellow line would represent the drop-per-mile calculated over 6,000 miles (as mentioned, about 3,500 feet per mile).
NOW can you see how it is impossible to describe a curve with a single gradient?
Scintific Method:
You have devolved into attacking me instead of the argument. Calling me ignorant (or any deviation from the discussion at hand) only proves you are lost and confused and can no longer provide any logical arguments.
Your sketch neither proves nor illustrates anything practical. You are playing the "fuzzy math" game. That is a fail.
This discussion of convexity/curvature/fall or drop on a globe is really quite simple. Let me help you out here...
Imagine selecting a point (Point A) on a large ball (say 10 feet in diameter) and turn the ball so that point A is on the top of the ball. Now imagine selecting another point (Point B) on that ball about 3 feet away - along the curvature of the ball, following the curvature of the ball. Now draw a horizontal line straight out, horizontally from Point A, past Point B.
Now simply measure from the point where your Horizontal line from Point A intersects with the Perpendicular line of Point B.
Measure the distance from the intersection of those lines (this is the Height of Point A) down to the dot of Point B. This is the amount of fall or drop from Point A to Point B. Hopefully you can now understand that there are NOT "multiple gradients" when calculating fall on a globe. It's very, very, very simple.
Go get yourself a ball of some sort and two rulers and you can hopefully understand "rate of fall" on a globe from one point to the next.
Best of luck with your studies.
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sceptimatic:
You said: "I don't agree with the calculations. It fails to take into account what I put in my diagram.
I agree that the earth cannot and is not a spinning ball but the calculations would not be 8 inches per mile and 8 more inches after that for each mile from a persons view point. It has to double for each mile."
No, I don't think so. How can the drop/fall possibly double with each mile travelled? No offence, but you need to re-think that one.
If the earth was a globe, which it isn't and you were to drive round it, then yes, every mile you covered you would assume the 8 inch curve. I understand that.
What I'm talking about is by viewing a building and moving back a mile at a time. Each mile moved, creates more curvature BOTH ways as the globe would be uniform, or basically, to save argument of the so called bulge.
Ok, I've re-evaluated. It would only double if two objects were moving away from each other by a mile EACH, not one object being stationary.
sceptimatic:
Right on man! You solved your own math problem, and you are using critical thinking! That is indicative of intelligence. It is also a great indicator that you say: "Oops, I was looking at that from a slightly different view."
I hope you take the time to look at my response to Scintific Method. I believe the "equation" is spelled on quite clearly there. Again, now matter where on a ball (representing Earth) you place a dot (Point A), it is actually the "top" or highest point of the ball. Any travel in any direction along the curvature of that ball is DOWN, and the intersection of the Horizontal "A" and the Perpendicular of "B" marks a measurable point.
I hope you can "picture" that. I'd love to have you argue WITH me rather than against me.
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Water can not possibly be anything other than flat and seeking it's own level at ALL times.
put a small drop of water on a flat surface.
Spank86:
Ha! That is an excellent argument except for one thing, we are not discussing a single drop of water. We are discussing oceans, lakes and rivers.
I could say in response to you: OK, pour one cup (a very small amount obviously) of water on a ball. Does it seek it's own level? Yes. Pour one cup of water on the slightest grade on earth. RET says that "gravity" will hold that water in place and it will not run down the slight gradient to seek it's own level.
WHY doesn't "gravity" hold that one cup of water on a slight gradient when it allegedly holds oceans of water in place with it's "magical" strength of attraction?
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Scintific Method:
You said: "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."
Wow, I'd really like to see water that does not lie flat, and water which is "forming a curve whichever way you look."
Water can not possibly be anything other than flat and seeking it's own level at ALL times.
It's not flat, it's curved, following the curve of the earth, and settling to the lowest points it can (aka the points closest to the centre of the earth, thanks to gravity); finding it's own level so that the surface of any single water mass forms a curve (albeit a very slight curve, not easily perceived with the unaided eye) with a pretty constant radius, the origin point for which is the centre of the earth.
Did you follow that? or do I need to dumb it down even more?
Scintific Method:
Yes, please dumb it down further for me.
How about showing me a pan of your curved water, OK? Just put it on YouTube and the whole world can watch and be awed.
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Imagine selecting a point (Point A) on a large ball (say 10 feet in diameter) and turn the ball so that point A is on the top of the ball. Now imagine selecting another point (Point B) on that ball about 3 feet away - along the curvature of the ball, following the curvature of the ball. Now draw a horizontal line straight out, horizontally from Point A, past Point B.
Now simply measure from the point where your Horizontal line from Point A intersects with the Perpendicular line of Point B.
Measure the distance from the intersection of those lines (this is the Height of Point A) down to the dot of Point B. This is the amount of fall or drop from Point A to Point B...
Now add a third point, point C, repeat the measurement from A to C, and you will get a different gradient. That's what I have been trying to explain.
Here's another diagram, following your suggestion and expanding on it:
(http://img689.imageshack.us/img689/7263/g8yq.png)
"O" is our origin point (your "Point A"), the distance OA is 1/16 the circumference of our globe, let's call this distance "x". This lets us put the other distances in terms of x, so OB is 2x, OC is 3x, and OD is 4x. "a", "b", "c", and "d" are the respective drops from the line drawn tangent to point "O", with a = 0.194x, b = 0.746x, c = 1.572x, and d = 2.546x. With me so far? I hope so! This stuff is not particularly complicated, and can be demonstrated easily using the method you suggested.
So, to finish off:
- gradient of OA is 0.194 (0.194 / 1);
- gradient of OB is 0.373 (0.746 / 2);
- gradient of OC is 0.524 (1.572 / 3);
- and the gradient of OD is 0.637 (do I need to show it? I guess so: 2.546 / 4).
NOW do you follow what I've been saying??
PS. If you were thinking "but those lines are not like what I was saying!", think about this: if i made a, b, c, and d perpendicular to the surface of the globe, they would be even longer (d would be infinite), and the gradients even more drastically different.
Oh, and for the sake of tidying this up a bit:
Scintific Method:
You said: "It is not "flat", but it does seek it's own level. To put this as clearly as I can, water will settle with it's surface being at a fairly constant radius from the centre of the earth, forming a curve whichever way you look."
Wow, I'd really like to see water that does not lie flat, and water which is "forming a curve whichever way you look."
Water can not possibly be anything other than flat and seeking it's own level at ALL times.
It's not flat, it's curved, following the curve of the earth, and settling to the lowest points it can (aka the points closest to the centre of the earth, thanks to gravity); finding it's own level so that the surface of any single water mass forms a curve (albeit a very slight curve, not easily perceived with the unaided eye) with a pretty constant radius, the origin point for which is the centre of the earth.
Did you follow that? or do I need to dumb it down even more?
Scintific Method:
Yes, please dumb it down further for me.
How about showing me a pan of your curved water, OK? Just put it on YouTube and the whole world can watch and be awed.
I'll go you one better. This was on YouTube:
The earth is round DERP (http://#)
Explain how that could happen on a flat earth.
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Not this dodgy video again!
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LightHouse Rosslare. Panasonic HC-V520 Zoom Test. (http://#ws)
Explain how this could happen on a round earth.
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scintific method:
Your video of the ship sailing away over the curvature of the Earth was disproved over 100 years ago. This is an optical illusion that is EASILY disproved with use of a telescope.
The ship the sailed "over the hill" was not falling off a slope, it was disappearing from naked eyesight and only APPEARED to sail off down-hill.
If the ship were at it's maximum distance and had seemed to disappear down the hill, it would return immediately in FULL VIEW if you looked at it through a telescope.
Did it pick up speed when it started going down hill? Did the Captain need to apply the brakes?
I think now would be a good time to read some introductory facts, FAQ's and critical documents to FET. For example if you had read "100 Proofs Earth Is Not A Globe", you would NEVER have tried to prove the Earth was a globe using a video whose premise was proven false before 1900 a.d.
tappet's video is PROOF the Earth can NOT be a globe. According to the long distance views through those telescopic lenses, there is NO convex angles or fall to the water.
tappet's video disproves your video.
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LightHouse Rosslare. Panasonic HC-V520 Zoom Test. (http://#ws)
Explain how this could happen on a round earth.
tappet:
Thank you! You saved me all the time it would have taken to find and post that video. It is perfect proof that The Earth Is Not A Globe!
I wish we knew the distances on those telescopic shots, then calculate curvature from the RET. It would be so simple to ask the RE believers: Where Is Your Globe? Where Is Your Convex Earth? Where Is Your "Fall"?
Thank's again tappet!
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LightHouse Rosslare. Panasonic HC-V520 Zoom Test. (http://#ws)
Explain how this could happen on a round earth.
With ease. As I pointed out before, due to the fact that the lighthouse and observer are both stationary, this video only serves to show how inability to see an object because of angular resolution can be overcome with magnification. It does not prove or disprove either a round or flat earth.
BTW, I notice you did not post your other video, the one of the yacht. Pity, as it shows something else which can only happen on a round earth: an object (perhaps I should call it a reference point) which, close up, appears well below the horizon, yet far away appears above the horizon, from the same vantage point. A physical impossibility on a flat earth.
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scintific method:
Your video of the ship sailing away over the curvature of the Earth was disproved over 100 years ago. This is an optical illusion that is EASILY disproved with use of a telescope.
The ship the sailed "over the hill" was not falling off a slope, it was disappearing from naked eyesight and only APPEARED to sail off down-hill.
If the ship were at it's maximum distance and had seemed to disappear down the hill, it would return immediately in FULL VIEW if you looked at it through a telescope.
Did it pick up speed when it started going down hill? Did the Captain need to apply the brakes?
I think now would be a good time to read some introductory facts, FAQ's and critical documents to FET. For example if you had read "100 Proofs Earth Is Not A Globe", you would NEVER have tried to prove the Earth was a globe using a video whose premise was proven false before 1900 a.d.
tappet's video is PROOF the Earth can NOT be a globe. According to the long distance views through those telescopic lenses, there is NO convex angles or fall to the water.
tappet's video disproves your video.
Your arguments serve only to display your ignorance. I have discussed Rowbotham's 'perspective' and "100 proofs..." previously, and the conclusion was that they were both worthless due to their inconsistencies and inaccuracies.
That video is quite clearly done using a substantial amount of magnification, so the claim that a telescope can restore the hull to view once it has been obscured by the horizon is clearly false. Please bring a better argument next time.
Oh, and I notice you have had nothing to say (yet) regarding my gradients explanation, why is that?
PS. please read my comment on tappet's video.
PPS. You still seem to be confused about which way gravity pulls an object, an opinion I have come to after you wrote this: "Did it pick up speed when it started going down hill? Did the Captain need to apply the brakes?" I suggest you give some more thought to the meaning of "toward the centre of the earth".
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LightHouse Rosslare. Panasonic HC-V520 Zoom Test. (http://#ws)
Explain how this could happen on a round earth.
With ease. As I pointed out before, due to the fact that the lighthouse and observer are both stationary, this video only serves to show how inability to see an object because of angular resolution can be overcome with magnification. It does not prove or disprove either a round or flat earth.
As I have asked before show me "sinking ship on flat and glassy sea with no swell " any surf board rider with years of experience would know ships are tiny dots on the horizon when its flat sea's and as the swell increases ships start sinking closer to shore, at this point you will notice you can easily see the ship with your eyes no magnification needed. The bigger the swell the closer the ships are sinking to the shore line. Your video has plenty of swell just check out the bobbing sea birds, check out the lumpy horizon. Why is the film constantly cut? , how high is the observer, is the observer stationary, to many variables and not "scintific" enough to be of any value.
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LightHouse Rosslare. Panasonic HC-V520 Zoom Test. (http://#ws)
Explain how this could happen on a round earth.
tappet:
Thank you! You saved me all the time it would have taken to find and post that video. It is perfect proof that The Earth Is Not A Globe!
I wish we knew the distances on those telescopic shots, then calculate curvature from the RET. It would be so simple to ask the RE believers: Where Is Your Globe? Where Is Your Convex Earth? Where Is Your "Fall"?
Thank's again tappet!
Sixstringthing you might enjoy these.
Panasonic SDR-H40.Zoom test Ship (http://#)
Nikon Coolpix P510 zoom test (ship at the sea) (http://#ws)
Sixstringthing, when you see a ship sinking on the horizon you do not need magnification they will be very close. When the sea is flat the ship will be so far away you cannot see it with the naked eye you need magnification to bring it into view.
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LightHouse Rosslare. Panasonic HC-V520 Zoom Test. (http://#ws)
Explain how this could happen on a round earth.
With ease. As I pointed out before, due to the fact that the lighthouse and observer are both stationary, this video only serves to show how inability to see an object because of angular resolution can be overcome with magnification. It does not prove or disprove either a round or flat earth.
As I have asked before show me "sinking ship on flat and glassy sea with no swell " any surf board rider with years of experience would know ships are tiny dots on the horizon when its flat sea's and as the swell increases ships start sinking closer to shore, at this point you will notice you can easily see the ship with your eyes no magnification needed. The bigger the swell the closer the ships are sinking to the shore line. Your video has plenty of swell just check out the bobbing sea birds, check out the lumpy horizon. Why is the film constantly cut? , how high is the observer, is the observer stationary, to many variables and not "scintific" enough to be of any value.
And as I responded, a 2-4ft swell will not hide a 50ft ship from an observer 20ft+ above the water level. I can understand how you would have the impression of the ships going out of sight closer when the sea is rougher; sitting on a surf board is a very low vantage point, from which it would be easy for a swell to hide a ship. Also a good place to take advantage of favourable refractive effects (such as experienced by Rowbotham at the Bedford Level) when the sea is calm. This has all been discussed elsewhere, but it is worth repeating.
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I feel I should have asked this earlier. In real life with your eyes have you seen a sinking ship?
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I feel I should have asked this earlier. In real life with your eyes have you seen a sinking ship?
Not a ship, but an oil (gas?) rig. On descent after a very enjoyable flight, I watched as it appeared to approach the horizon (even though my line-of-sight distance to it never changed), 'crest' the horizon, and then sink out of sight behind it. Perhaps a better way to word that would be to say that the horizon appeared to come closer as I descended.
If you can convince me, leaving no doubt in my mind, that this is possible on a flat earth, you might just earn a convert! (maybe ;) )
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I feel I should have asked this earlier. In real life with your eyes have you seen a sinking ship?
Not a ship, but an oil (gas?) rig. On descent after a very enjoyable flight, I watched as it appeared to approach the horizon (even though my line-of-sight distance to it never changed), 'crest' the horizon, and then sink out of sight behind it. Perhaps a better way to word that would be to say that the horizon appeared to come closer as I descended.
If you can convince me, leaving no doubt in my mind, that this is possible on a flat earth, you might just earn a convert! (maybe ;) )
Thanks for being honest.
You misunderstand me. I am not trying to convince you the earth is flat, I honestly have no idea what shape this earth is. I only lurk on this site to learn from flat and round earthers. But if you can find somebody you trust that has seen sinking ship ask them if they needed a scope or just used their eyes. My guess is they will say just with their naked eyes because sinking ships are very easy to see. Once this has been established ask yourself how it could be possible to need to zoom the ships or lighthouse in the video's I posted.
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Apologies for the misunderstanding! I came to this site to see if there was a credible alternative explanation for many of the things that a round earth would account for, such as the 'sinking ship effect'. So far there has been nothing convincing. I have also conducted a number of experiments and observations of my own, and given much thought to the possibility of a flat earth. I even started to formulate my own "flat earth theory"! All that I have achieved in all of this is to become more certain that what we are told is true: the earth is round, not perfectly round, but very close to it.
Go to the lighthouse video, pause it at 9 seconds, then again at 10 seconds. At 9 seconds, the lighthouse is not visible. At 10 seconds, it is visible, 1 pixel wide and about 3-4 pixels high (at the resolution of the posted video, and at the point I paused). If you keep playing until the lighthouse is well zoomed in, you'll see that the lighthouse is about 4 times higher than it is wide. What does this tell me? That the lighthouse has not been obscured in any way, it's just too small to see without magnification. Before the 10 second mark, the lighthouse is less than 1 pixel wide, so it doesn't show. This is equivalent to not being able to see a single human hair beyond a few metres: the angular resolution is less than the human-perceivable minimum at that distance. Does this make sense?
Keeping that in mind, pop back to the video of the tall ship going over the horizon. It only disappears from view after most of it has already been obscured by the horizon. What does this tell you? Please, also keep in mind that the prevailing swell would be physically incapable of hiding any more than a few boards of the hull, just above the waterline, given the vantage point of the observer.
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I don't think that you are understanding how close and how big sinking ships are when you see them. It makes no difference whether you stand in the water or are 50ft above they still look big. When a container ship appears to sink you can make out a single container with clarity using only your eyes, its possible to see only the bridge and a container on the bow with clarity but no hull whatsoever with no magnification or zoom needed. So if you see a sinking ship that is what you will see.
Remember you do not need zoom to see sinking ship it is always way to big to need it.
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I don't think that you are understanding how close and how big sinking ships are when you see them. It makes no difference whether you stand in the water or are 50ft above they still look big. When a container ship appears to sink you can make out a single container with clarity using only your eyes, its possible to see only the bridge and a container on the bow with clarity but no hull whatsoever with no magnification or zoom needed. So if you see a sinking ship that is what you will see.
Remember you do not need zoom to see sinking ship it is always way to big to need it.
Okay, I have never seen one that close! Please though, keep in mind, when sitting on a surfboard on the water, a 4ft swell will be higher than your eye line and will easily block your view, but when you are 20ft above the water, a 4ft swell will block no more than 4ft of what is behind it, most of the time much less.
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Water can not possibly be anything other than flat and seeking it's own level at ALL times.
put a small drop of water on a flat surface.
Spank86:
Ha! That is an excellent argument except for one thing, we are not discussing a single drop of water. We are discussing oceans, lakes and rivers.
I could say in response to you: OK, pour one cup (a very small amount obviously) of water on a ball. Does it seek it's own level? Yes. Pour one cup of water on the slightest grade on earth. RET says that "gravity" will hold that water in place and it will not run down the slight gradient to seek it's own level.
WHY doesn't "gravity" hold that one cup of water on a slight gradient when it allegedly holds oceans of water in place with it's "magical" strength of attraction?
Gravity says nothing of the sort, it will seek it's own level but that level is based on what you perceive as flat actually being curved over an incredibly large distance. Gravity pulls towards the center of the earth at all times, A basketball doesn't have enough mass to hold water to it and the experiment is even more flawed when conducted within an even greater gravity field (on the earth).
I think with regard to gradients what the others are saying is that the rate of fall on a curve depends greatly on the distance you're measuring. So measuring over 1 mile will give a lesser rate of drop than over 2 miles, than over 3... and so on and so forth. This is because the curve is curving away all the time.
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Scintific Mind: You said:
"Please, also keep in mind that the prevailing swell would be physically incapable of hiding any more than a few boards of the hull, just above the waterline, given the vantage point of the observer."
Forget your swells arguments and other complications of a simple argument. They are IRRELEVANT. This is a simple truth we seek.
HERE IS THE SOLUTION TO THE SHIPS SAILING OVER THE HORIZON AND DROPPING:
Let us take a LOCOMOTIVE train, one big unit. Let's run it over Flat Earth, say the "Bonneville Salt Flats" or the state of Kansas in the USA.
Keep in mind that railroad tracks are made perfectly straight, and when they "lay" tracks over any long FLAT distance, anywhere on Earth, the engineers NEVER EVER COMPENSATE FOR CURVATURE OF THE EARTH.
THAT IS BECAUSE THERE IS NO CURVATURE OF THE EARTH.
The same reason Engineers NEVER EVER compensate for curvature of the Earth when making long tunnels and canals.
Watch the train disappear or "drop over the horizon" (as you want to imagine), on perfectly engineered railroad tracks and please, tell me sir... how do you account for this? Is there magically appearing CURVATURE In the steel tracks?
Scintific Method: You should call all of the Railroad Engineers, Canal Engineers and Tunnel Engineers and tell them they are making a HUGE MISTAKE!
Tell them they have somehow forgotten to take into consideration the CURVATURE of the Earth when they Engineered and built their incredibly FLAT engineering marvels! They should never design anything FLAT on a Round Earth!
Tell them to go back and have another look at their mathematics and calculations and find the error of their ways!
THERE IS NO WAY THE EARTH IS A GLOBE.
Please stop trying to justify that NONSENSE with outdated arguments about "swells" and ships falling over the horizon, and water NOT seeking it's own level, and all the other nonsense you are trying to peddle.
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How can you say that rails are perfectly straight? That's a ridiculous claim! There is more undulation in 100m of track than there is curvature, because the rate of curvature is less than 0.0009° over that distance! Again, your ignorance is showing through.
EDIT: I may have been a little overzealous in my initial posting; there is, in fact, a 3km long section of absolutely, perfectly straight track near where I live. If you were to measure it with a laser (done), there would be almost no deviation from a straight line at all. The interesting thing is, if you place a spirit level at either end (done), it will indicate a gentle slope toward the middle! I doubt you would understand the reason for that, but please, make an effort to.
How long is the longest tunnel on earth? (154.3km) Is it perfectly straight? (No) Another useless argument that proves nothing!
What is the datum that canal engineers use? (sea level, which follows the curve of the earth) How long is any single surveyed segment of the canal? (this is just a guess, but my passing familiarity with surveying equipment suggests the limit is 500m, over which distance the error created by the earth's curve is only 2cm, or 0.004°) Another useless argument that proves nothing!
Improve your knowledge of round earth physics before attacking it, because you are looking awfully foolish right now.
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How can you say that rails are perfectly straight? That's a ridiculous claim! There is more undulation in 100m of track than there is curvature, because the rate of curvature is less than 0.0009° over that distance! Again, your ignorance is showing through.
EDIT: I may have been a little overzealous in my initial posting; there is, in fact, a 3km long section of absolutely, perfectly straight track near where I live. If you were to measure it with a laser (done), there would be almost no deviation from a straight line at all. The interesting thing is, if you place a spirit level at either end (done), it will indicate a gentle slope toward the middle! I doubt you would understand the reason for that, but please, make an effort to.
How long is the longest tunnel on earth? (154.3km) Is it perfectly straight? (No) Another useless argument that proves nothing!
What is the datum that canal engineers use? (sea level, which follows the curve of the earth) How long is any single surveyed segment of the canal? (this is just a guess, but my passing familiarity with surveying equipment suggests the limit is 500m, over which distance the error created by the earth's curve is only 2cm, or 0.004°) Another useless argument that proves nothing!
Improve your knowledge of round earth physics before attacking it, because you are looking awfully foolish right now.
Perfect! You say 154 Km tunnel is not perfectly straight? (Once AGAIN you confuse the question with fuzzy BS that means absolutely nothing. You are becoming adept at obfuscation.) "Is the tunnel straight?" - is NOT the question! WHERE IS THE CURVATURE/FALL CALCULATIONS that would be required by the Engineers to keep the tunnel level over 154 Km?
You insist on calling me ignorant (name calling: obvious lack of subject knowledge) while you PERSIST with trying to deviate from simple, straightforward uncomplicated questions with drawings and cartoons and bullsh*t "data" that has absolutely no value to the consideration at hand.
You must be paid to try to derail the momentum of Flat Earth truth. Why else would someone argue so bitterly using illogical nonsense?
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Perfect! You say 154 Km tunnel is not perfectly straight? (Once AGAIN you confuse the question with fuzzy BS that means absolutely nothing. You are becoming adept at obfuscation.) "Is the tunnel straight?" - is NOT the question! WHERE IS THE CURVATURE/FALL CALCULATIONS that would be required by the Engineers to keep the tunnel level over 154 Km?
For a start the tunnel isn't level and doesn't need to be level.
For second If the earth curves due to gravity, and water curves for the same reason then everything made straight within that gravity will curve too (assuming it's made with a spirit level). Plus any height measurements would be taken from sea level... A sea level that would be curved on a round earth so boring a tunnel from both ends would still meet despite the curvature even if one end was longer than the other.
The simple facts are that this can't prove a round or a curved earth by itself although a comparison between a laser level and a spirit level over a long distance could (or could at least prove something odd was going on) if you could get the tolerances close enough.
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Perfect! You say 154 Km tunnel is not perfectly straight? (Once AGAIN you confuse the question with fuzzy BS that means absolutely nothing. You are becoming adept at obfuscation.) "Is the tunnel straight?" - is NOT the question! WHERE IS THE CURVATURE/FALL CALCULATIONS that would be required by the Engineers to keep the tunnel level over 154 Km?
For a start the tunnel isn't level and doesn't need to be level so there would be no need for such calculations whatever the shape of the earth.
For second If the earth curves due to gravity, and water curves for the same reason then everything made straight within that gravity will curve too (assuming it's made with a spirit level). Plus any height measurements would be taken from sea level... A sea level that would be curved on a round earth so boring a tunnel from both ends would still meet despite the curvature even if one end was longer than the other.
The simple facts are that this can't prove a round or a curved earth by itself although a comparison between a laser level and a spirit level over a long distance could (or could at least prove something odd was going on) if you could get the tolerances close enough.
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Perfect! You say 154 Km tunnel is not perfectly straight? (Once AGAIN you confuse the question with fuzzy BS that means absolutely nothing. You are becoming adept at obfuscation.) "Is the tunnel straight?" - is NOT the question! WHERE IS THE CURVATURE/FALL CALCULATIONS that would be required by the Engineers to keep the tunnel level over 154 Km?
For a start the tunnel isn't level and doesn't need to be level.
For second If the earth curves due to gravity, and water curves for the same reason then everything made straight within that gravity will curve too (assuming it's made with a spirit level). Plus any height measurements would be taken from sea level... A sea level that would be curved on a round earth so boring a tunnel from both ends would still meet despite the curvature even if one end was longer than the other.
The simple facts are that this can't prove a round or a curved earth by itself although a comparison between a laser level and a spirit level over a long distance could (or could at least prove something odd was going on) if you could get the tolerances close enough.
Engineers don't calculate long spans of ANY type to include curvature of the Earth. That's because there is none.
If there WERE curvature of the Earth (think of the ship sailing away), Engineers would have to design a hell of a Drop into any Railroad track that was the length as the same distance as the distance away that ship is when it "sails over the horizon".
FOR EXAMPLE: Let's say that ship disappears at 7 miles out (whatever the distance is, doesn't matter for this analogy), and Engineers build a Railroad track 7 miles long.
Let's say for example that the ship is 70' tall (top of mast to bottom of hull).
That means that the Railroad track would be 70' lower when it is 7 miles away, just like the ship which sailed "over the horizon on the curvature of the Earth").
HOW would the Railroad Engineers design 70' of fall into this stretch of track when they are using (virtually straight) steel Railroad tracks?
Do you say the tracks aren't perfectly straight from the steel mill? OK, how much are they off? It must be an incredibly small amount.
The Earth is NOT a globe. It's just simple logic.
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Do you say the tracks aren't perfectly straight from the steel mill? OK, how much are they off? It must be an incredibly small amount.
They are off by exactly the curvature off the earth of course. That's the very point of Round earth theory.
Gravity pulls to the center of the round earth so when anything made with a spirit level it minutely follows the curvature.
Plus individual rails wouldn't need to be curved anyway, think of the joins between the rail, what are the tolerances there? easily enough to allow a curvature.
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FOR EXAMPLE: Let's say that ship disappears at 7 miles out (whatever the distance is, doesn't matter for this analogy), and Engineers build a Railroad track 7 miles long.
It kinda does matter if your entire argument is to show that the curvature isn't small enough to be difficult to notice.
I still don't understand why you think railroad engineers ever bother about making flat tracks anyway, they try to grade hills and bore relatively straight but it's just not important to them if stuff goes up and down a bit and if you bore from two ends at once well you're both off by the same amount.
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Do you say the tracks aren't perfectly straight from the steel mill? OK, how much are they off? It must be an incredibly small amount.
They are off by exactly the curvature off the earth of course. That's the very point of Round earth theory.
Gravity pulls to the center of the round earth so when anything made with a spirit level it minutely follows the curvature.
Plus individual rails wouldn't need to be curved anyway, think of the joins between the rail, what are the tolerances there? easily enough to allow a curvature.
Spank:
So you are saying that "gravity" pulls the steel railroad tracks onto the curvature of the Earth and if that doesn't work then the tolerances in the joints will allow for 70' or fall?
If the Earth was shaped like a Basketball and you laid 100 miles of virtually straight steel Railroad Track out, the track would stick up in the AIR after a few miles!
I do NOT think "gravity" is going to pull the steel Railroad track down to conform with the "curvature" of the "Basketball", do you?
I do NOT think the critical tolerances in the joints of Railroad tracks are enough to keep the Railroad track from sticking up in the air after a few miles traveling along the curvature of the "Basketball".
So you think virtually perfectly straight Railroad tracks will be pulled down to meet the curvature of the "Basketball".
Wow, amazing.
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sixstringthing, I call you ignorant because everything you say displays ignorance of the subject being discussed. I have tried my best to rectify this ignorance, by providing diagrams, examples, and explanations, in as clear a format as I can manage, but have been met by yet more ignorance. You clearly do not understand the subject at hand, and seem also to have no desire to. For this reason, I pity you. That said, I have not given up yet! Please, read on, and try to pay attention this time!
So you are saying that "gravity" pulls the steel railroad tracks onto the curvature of the Earth and if that doesn't work then the tolerances in the joints will allow for 70' or fall?
That's exactly what happens. The required amount of flex for a rail to match the curve of the earth is minimal. Do the engineers use bent sections of track to go over a hill? No, because the track flexes enough to follow the curve of the hill (which, you must admit, is far greater than that of the earth). Do they use bent sections around corners? No, because the track has enough flexibility in it to simply be bent as it is laid.
If the Earth was shaped like a Basketball and you laid 100 miles of virtually straight steel Railroad Track out, the track would stick up in the AIR after a few miles!
You must have missed my remark about the 3km section of perfectly straight track near my home. To refresh you, a spirit level placed on the track at either end indicates a gentle slop toward the middle, yet the track is exactly straight! Please, give some thought to the reason for that.
Engineers don't calculate long spans of ANY type to include curvature of the Earth. That's because there is none.
Really? Never? Can you prove that there has never been a long man-made structure that included calculations of curvature?
If there WERE curvature of the Earth (think of the ship sailing away), Engineers would have to design a hell of a Drop into any Railroad track that was the length as the same distance as the distance away that ship is when it "sails over the horizon".
No. Please read the above comments on railroads. And for pity's sake, learn something about the subject, and give it some careful thought, before trying to argue it!
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Scintific Method:
You said: "Really? Never? Can you prove that there has never been a long man-made structure that included calculations of curvature?"
When you get nowhere by name calling you get down to your last, most desperate hope at smearing the obvious with your fanciful imagery. You want me to prove a negative?
You want ME to prove there has never been something? Anything? LOL. Silly boy. Don't you know it's the act of desperation to ask your opponent to prove a negative?
UNFORTUNATELY FOR YOU, that turns the tables!
It is YOU who MUST prove a positive (or admit defeat... which you will!).
So... PROVE there HAS been a man-made structure that took into account the "curvature" of the Basketball (Earth).
Go ahead sir, prove it. Find one. JUST ONE! Of all the miracles created by man and huge distances spanned by tunnels, bridges, Railroads, Highways and Canals across the entire world, please fine me JUST ONE that took into consideration the "curvature" of your Basketball-shaped Earth.
When you can NOT find JUST ONE example, please return immediate with an appropriate apology for your confusion over such simple matters.
If, however, you DO find JUST ONE EXAMPLE of an Engineering Calculation (of any type structure, anywhere on Earth) that takes into consideration the "curvature" of your Basketball-shaped Earth, I will glad apologize for the "ignorance" you have repeatedly claimed I have.
I eagerly await your response:
PROVE IT!!!
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There have been countless, Sixstring.
None of them are on the scale you think of, because there's no structure with a singular component long enough and also inflexible enough to require being machined to the curvature of the earth.
But there are quite a number of structures that require the curvature to function.
They whiz over us all the time. Wouldn't work without the curvature, you see.
But for a terrestrial example, the Danyang–Kunshan Grand Bridge in China, a 102 mile bridge, Bang Na expressway in Thailand, at 34 miles, and the Lake Pontchartrain Causeway in southern Louisiana, which is 23.83 miles entirely over water. All have to take the curvature of the earth into account, not for individual pieces, but for how they join together to form a coherent whole.
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There have been countless, Sixstring.
None of them are on the scale you think of, because there's no structure with a singular component long enough and also inflexible enough to require being machined to the curvature of the earth.
But there are quite a number of structures that require the curvature to function.
They whiz over us all the time. Wouldn't work without the curvature, you see.
But for a terrestrial example, the Danyang–Kunshan Grand Bridge in China, a 102 mile bridge, Bang Na expressway in Thailand, at 34 miles, and the Lake Pontchartrain Causeway in southern Louisiana, which is 23.83 miles entirely over water. All have to take the curvature of the earth into account, not for individual pieces, but for how they join together to form a coherent whole.
[/quot
Countless? Awesome!
It should be an exceptionally easy task to show me JUST ONE example of the mathematics or Engineering plans from JUST ONE of these "countless" Engineering marvels that require Calculations for negotiating the "curvature" of the Basketball-shaped Earth.
I CAN'T WAIT TO SEE YOUR EXAMPLE!
PLEASE DO HURRY BACK WITH IT.
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I was indirectly asking you to back up this statement:
Engineers don't calculate long spans of ANY type to include curvature of the Earth.
Since you seem to want to put the onus on me to disprove your statement instead, here you go:
The Australia Telescope Compact Array is made up of multiple radio telescopes. These require a suitable and consistent reference point for synchronised observations, thus the track they run on had to be constructed in such a way that it would be perfectly straight for it's entire 3km length. Due to the curvature of the earth, this required that each end of the 3km track go "uphill" relative to the local pull of gravity, in order to maintain an exactly straight line from one end to the other. How do I know about this? I grew up virtually right next to it, and have had the opportunity to see this remarkable piece of engineering with my own eyes.
Is that good enough, or do you need to see the plans? They might be a little hard to get hold of, I honestly wouldn't know where to start looking...
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I was indirectly asking you to back up this statement:
Engineers don't calculate long spans of ANY type to include curvature of the Earth.
Since you seem to want to put the onus on me to disprove your statement instead, here you go:
The Australia Telescope Compact Array is made up of multiple radio telescopes. These require a suitable and consistent reference point for synchronised observations, thus the track they run on had to be constructed in such a way that it would be perfectly straight for it's entire 3km length. Due to the curvature of the earth, this required that each end of the 3km track go "uphill" relative to the local pull of gravity, in order to maintain an exactly straight line from one end to the other. How do I know about this? I grew up virtually right next to it, and have had the opportunity to see this remarkable piece of engineering with my own eyes.
Is that good enough, or do you need to see the plans? They might be a little hard to get hold of, I honestly wouldn't know where to start looking...
Please cite the source of your reference so that I may have a look. Thank you.
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I was indirectly asking you to back up this statement:
Engineers don't calculate long spans of ANY type to include curvature of the Earth.
Since you seem to want to put the onus on me to disprove your statement instead, here you go:
The Australia Telescope Compact Array is made up of multiple radio telescopes. These require a suitable and consistent reference point for synchronised observations, thus the track they run on had to be constructed in such a way that it would be perfectly straight for it's entire 3km length. Due to the curvature of the earth, this required that each end of the 3km track go "uphill" relative to the local pull of gravity, in order to maintain an exactly straight line from one end to the other. How do I know about this? I grew up virtually right next to it, and have had the opportunity to see this remarkable piece of engineering with my own eyes.
Is that good enough, or do you need to see the plans? They might be a little hard to get hold of, I honestly wouldn't know where to start looking...
Some construction plan would have to be filed with the local municipality or county maybe?
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I was indirectly asking you to back up this statement:
Engineers don't calculate long spans of ANY type to include curvature of the Earth.
Since you seem to want to put the onus on me to disprove your statement instead, here you go:
The Australia Telescope Compact Array is made up of multiple radio telescopes. These require a suitable and consistent reference point for synchronised observations, thus the track they run on had to be constructed in such a way that it would be perfectly straight for it's entire 3km length. Due to the curvature of the earth, this required that each end of the 3km track go "uphill" relative to the local pull of gravity, in order to maintain an exactly straight line from one end to the other. How do I know about this? I grew up virtually right next to it, and have had the opportunity to see this remarkable piece of engineering with my own eyes.
Is that good enough, or do you need to see the plans? They might be a little hard to get hold of, I honestly wouldn't know where to start looking...
Where did you get the following quote? Please cite this source.
"The Australia Telescope Compact Array is made up of multiple radio telescopes. These require a suitable and consistent reference point for synchronised observations, thus the track they run on had to be constructed in such a way that it would be perfectly straight for it's entire 3km length. Due to the curvature of the earth, this required that each end of the 3km track go "uphill" relative to the local pull of gravity, in order to maintain an exactly straight line from one end to the other."
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Where is your source?
You just quoted this statement, the source must be close at hand.
Please post the source.
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That was not a quote, it was acquired knowledge put into my own words. I do wish you would read my posts properly, I would have thought the following line would have given it away:
How do I know about this? I grew up virtually right next to it, and have had the opportunity to see this remarkable piece of engineering with my own eyes.
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Your first question about what direction I think gravity pulls in a RET situation? We are taught that it pulls towards the center of the "globe", or in the general direction of down. I'm not sure why you asked that or why it's relevant to my question?
It's relevant to the question it was in reply to, as it explains how water can be held in a "fixed curvature", as you put it.
Your answer as to the "rate" or curvature of the "globe/Earth":
"Giving the curvature in inches per mile can be misleading. If I do the calculation over 1 mile, I get 8.6 inches (close enough to your quoted value). If, however, I do it over, say, 100 miles, I get 71.6 feet per mile. You can do these calculations yourself easily enough if you understand trigonometry."
OK... a drop of 71.6 feet over 100 miles IS 8.6 inches per mile.
I guess unless we hear from anyone else, the "rate" is 8 - 8.6 inches per mile.
With THAT said it is incontrovertible proof that there is NO curvature to the Earth based on past experiments measuring "rate of fall" (or convexity) of Earth over water. Such experiments have been discussed countless times and stated in different ways. NO "rate of fall" can be calculated anywhere over water.
Conclusion: The Earth is Flat, there can be no other explanation
Pay attention. I didn't say "71.6 over 100 miles" I said "71.6 feet per mile". Over 100 miles it would be ~7,160 feet. This can be confirmed by going for a fly (which I have done countless times in my life), and noting that, when at 7,000 feet, you can see about 100 miles. Standing on the ground, you can see for about 3 miles. From about 100 feet above the ground, you can see roughly 12 miles. I've personally verified all of this, and I consider it to be pretty sound proof of curvature. I encourage you to go and verify it for yourself as well, and/or come up with a credible alternative explanation for it.
EDIT: No, I didn't stuff up my calculation, the fact that 71.6 feet divided by 100 is about 8.6 inches is just a coincidence. It threw me off when I first did the maths, so I triple checked it, and it is most definitely right. As an added note, the drop over 6,000 miles is 3,981 miles, or about 3500 feet per mile. If you can tell me why that is, I'll refrain from calling you names. ;)
I served in the Navy for 4 years, but to tell the truth, I never was much of a true sailor. :'( Like a lot of other persons who were specialists in their own specialty ratings , I was more of a technician. I never did get interested is such things as navigation, etc. Maybe I should have been.
But all of that aside, the curvature of the so-called "Round Earth" or more specifically a spherical earth was apparent from my observations. I do know from personal experience that I could see much farther to the horizon if I was high on the 04 Deck of the ship than when I was on the Liberty Boat at sea level. Height of the observer has to be taken into account when determing the distance to the horizon, which proves the curvature of the earth. (That was the reason for those "Crow's Nests" being placed so high on those masts on those old ships.) Of course this could also be said for the Flat Earth, but no curvature would have to be taken into account in that case.
A Theoretical Question for the Flat Earth Experts.:
Theoretically speaking, if you could go high enough above the earth, and if the earth was flat, why couldn't you see the whole Flat Earth as a disc and see the Ice Ring , etc. ? (This is based on the Unipolar Map of the Flat Earth. But this poses a problem.)One post said "There is no Flat Earth Map." And you also couldn't go high enough since Flat Earthers also say "There is no space flight" and you couldn't get high enough.
Also, why if you go high enough, you can only see half of the earth ? (A "Round Earth" that is.) OOOOPS !!!!!! Sorry about that one. Flat Earthers also say those photographs are all fakes and just drawn to show a spherically shaped earth.
P.S. Don't ever say the words "Flat Earth" to a sailor or an ex-sailor. You might find their reply to be highly embarrassing to you. If the earth was flat why couldn't I see Honolulu or San Diego much farther out at sea than I could ?. Oh, right, the old "perspective". But there were such things as binoculars and telescopes. They didn't work on those "sinking ships" also.
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Go ahead sir, prove it. Find one. JUST ONE! Of all the miracles created by man and huge distances spanned by tunnels, bridges, Railroads, Highways and Canals across the entire world, please fine me JUST ONE that took into consideration the "curvature" of your Basketball-shaped Earth.
anything using a spirit level.
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That was not a quote, it was acquired knowledge put into my own words. I do wish you would read my posts properly, I would have thought the following line would have given it away:
How do I know about this? I grew up virtually right next to it, and have had the opportunity to see this remarkable piece of engineering with my own eyes.
Well, it sure looked and sounded like a quote when it was made in response to my request for proof of ANY Engineered structure that used the "curvature" of Earth in it's mathematical or Engineering Construction calculations.
OK, so you made that up. Fine.
Now, go back to my original response and please find one, JUST ONE, man-made structure of ANY TYPE that uses "curvature" of the Basketball-shaped Earth in it's design, mathematical or Engineering calculations.
I happen to know that you will not find such evidence because there are NO structures that take into consideration the "curvature" of the Earth.
If there was a Spherical Earth, it would be ABSOLUTELY NECESSARY to calculate for the fall related to the curvature when constructing/Engineering long projects.
But since the EARTH IS NOT A GLOBE, you will not find ONE SINGLE EXAMPLE of an engineered structure with Engineering calculations related to the "curvature" of the Earth.
But I'll be waiting, just in case you find one, JUST ONE, example.
Or, you could just admit that defending Round Earth Theory is impossible because EARTH IS NOT A GLOBE.
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But since the EARTH IS NOT A GLOBE, you will not find ONE SINGLE EXAMPLE of an engineered structure with Engineering calculations related to the "curvature" of the Earth.
I have a question... why?
I mean all calculations between two point are made by the engineers based on a round earth so you're not going to drop short due to curvature and the only really long constructions are tunnels and bridges and they both have tolerance built in anyway and meander up and down far more than round earth curvature.
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But since the EARTH IS NOT A GLOBE, you will not find ONE SINGLE EXAMPLE of an engineered structure with Engineering calculations related to the "curvature" of the Earth.
I have a question... why?
I mean all calculations between two point are made by the engineers based on a round earth so you're not going to drop short due to curvature and the only really long constructions are tunnels and bridges and they both have tolerance built in anyway and meander up and down far more than round earth curvature.
I don't think calculations between two (distant) points ARE made by engineers based on a round Earth.
As a matter of fact, we are trying to fine ONE, JUST ONE, example where calculations ARE made based on the curvature of Earth.
However, it seems no one can find one, JUST ONE, example of any engineered structure that calculated based on the curvature of the Earth.
Therefore, Engineers apparently do NOT make calculations based on the curvature of Earth, or a Round Earth, but they apparently make calculations based on a Flat Earth.
So now you know.
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But since the EARTH IS NOT A GLOBE, you will not find ONE SINGLE EXAMPLE of an engineered structure with Engineering calculations related to the "curvature" of the Earth.
I have a question... why?
I mean all calculations between two point are made by the engineers based on a round earth so you're not going to drop short due to curvature and the only really long constructions are tunnels and bridges and they both have tolerance built in anyway and meander up and down far more than round earth curvature.
Yes. If you use a round Earth calculation and expect a round Earth result, you get one. Congratulations.
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OK, so you made that up. Fine.
So I made that up? Prove it! I know with certainty that you can't, because I didn't!
Now, go back to my original response and please find one, JUST ONE, man-made structure of ANY TYPE that uses "curvature" of the Basketball-shaped Earth in it's design, mathematical or Engineering calculations.
I happen to know that you will not find such evidence because there are NO structures that take into consideration the "curvature" of the Earth.
There you go again, making big claims without being able to back them up (have you actually gone through the plans for every structure ever built, looking for these calculations?), and ignoring the examples that have already been given (thanks, hewholikespie!). And you wonder why I call you ignorant!!
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OK, so you made that up. Fine.
So I made that up? Prove it! I know with certainty that you can't, because I didn't!
Now, go back to my original response and please find one, JUST ONE, man-made structure of ANY TYPE that uses "curvature" of the Basketball-shaped Earth in it's design, mathematical or Engineering calculations.
I happen to know that you will not find such evidence because there are NO structures that take into consideration the "curvature" of the Earth.
There you go again, making big claims without being able to back them up (have you actually gone through the plans for every structure ever built, looking for these calculations?), and ignoring the examples that have already been given (thanks, hewholikespie!). And you wonder why I call you ignorant!!
Chill out. If you didn't make it up, where did it come from?
Remember, it's impossible to prove a negative. YOU say there are countless projects that use curvature in their Engineering calculations. Well, where IS one?
I can't dig through construction plans looking for non-existing calculations, but YOU, since you claim there are so many projects that use "curvature" calculations in project engineering, should just go get ONE to show us.
Good luck with that.
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But since the EARTH IS NOT A GLOBE, you will not find ONE SINGLE EXAMPLE of an engineered structure with Engineering calculations related to the "curvature" of the Earth.
I have a question... why?
I mean all calculations between two point are made by the engineers based on a round earth so you're not going to drop short due to curvature and the only really long constructions are tunnels and bridges and they both have tolerance built in anyway and meander up and down far more than round earth curvature.
Yes. If you use a round Earth calculation and expect a round Earth result, you get one. Congratulations.
jroa, I sure hope you're here to defend truth, justice and Flat Earth. The RE Theorists are trying to gang up on me. Not that they're having any success, mind you.
Arguing with a RE Theorists is like dueling with a blind man. Not a lot of competition.
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I am here to defend truth.
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I am here to defend truth.
What do you think of my arguments so far? (remember, I'm a brand-new FEer)
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I don't think calculations between two (distant) points ARE made by engineers based on a round Earth.
I think you missed my point. Assuming a hypothetical round earth... why would they need to be?
As a matter of fact, we are trying to fine ONE, JUST ONE, example where calculations ARE made based on the curvature of Earth.
And I'm saying it would be highly unlikely to be necessary since the only long distance structures are relatively independent of round or flat earths. they work either way. the only ones I can think of that might not be are like the one scientific stated.
Therefore, Engineers apparently do NOT make calculations based on the curvature of Earth, or a Round Earth, but they apparently make calculations based on a Flat Earth.
So now you know.
so you think all engineers believe in a flat earth and somehow neglected to mention?
But since the EARTH IS NOT A GLOBE, you will not find ONE SINGLE EXAMPLE of an engineered structure with Engineering calculations related to the "curvature" of the Earth.
I have a question... why?
I mean all calculations between two point are made by the engineers based on a round earth so you're not going to drop short due to curvature and the only really long constructions are tunnels and bridges and they both have tolerance built in anyway and meander up and down far more than round earth curvature.
Yes. If you use a round Earth calculation and expect a round Earth result, you get one. Congratulations.
so do you think the engineers all believe in a flat earth and are using flat earth calculations?
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Chill out. If you didn't make it up, where did it come from?
No need to chill, I'm highly amused! I got that information from personal experience and observation, so it's a little hard to provide a citation.
YOU say there are countless projects that use curvature in their Engineering calculations.
Nope, not me. I have no doubt there would be a few out there that actually had to take it into account (hence my response to your (negative, and therefore unprovable and unsound) claim that there were none), but I leave it to others to find them. I have a perfect example of a structure that compensates for the earth's curve only a few k's from where I live now, and a number of other observations not related to the current topic that prove beyond doubt that the earth is round.
I can't dig through construction plans looking for non-existing calculations, but YOU, since you claim there are so many projects that use "curvature" calculations in project engineering, should just go get ONE to show us.
Good luck with that.
Again, not my claim, I leave the legwork on that one to those who made it.
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Where is your source?
You just quoted this statement, the source must be close at hand.
Please post the source.
"The whole basis of an array is that only the signal being tracked is common to every antenna. By correlating all the data from all the antennas, the common signal can be separated out. All of the project's key components were designed and manufactured within the Division of Radiophysics or by Australian industry. The telescope's engineering manager John Brooks, said the project required a level of precision and quality control probably unprecedented in Australian civil engineering. For example, the 3-kilometre-wide gauge rail track laid on thousands of tonnes of granite ballast had to be accurate to within plus or minus 0.5 millimetres from end-to-end while accommodating the curvature of the Earth."
Brad collis.
from here, search for "curv"
http://www.csiropedia.csiro.au/display/CSIROpedia/Australia+Telescope+Compact+Array (http://www.csiropedia.csiro.au/display/CSIROpedia/Australia+Telescope+Compact+Array)
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Thanks Spank, I actually was looking for that!
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Scintific Method:
You said: "Really? Never? Can you prove that there has never been a long man-made structure that included calculations of curvature?"
When you get nowhere by name calling you get down to your last, most desperate hope at smearing the obvious with your fanciful imagery. You want me to prove a negative?
You want ME to prove there has never been something? Anything? LOL. Silly boy. Don't you know it's the act of desperation to ask your opponent to prove a negative?
UNFORTUNATELY FOR YOU, that turns the tables!
It is YOU who MUST prove a positive (or admit defeat... which you will!).
So... PROVE there HAS been a man-made structure that took into account the "curvature" of the Basketball (Earth).
Go ahead sir, prove it. Find one. JUST ONE! Of all the miracles created by man and huge distances spanned by tunnels, bridges, Railroads, Highways and Canals across the entire world, please fine me JUST ONE that took into consideration the "curvature" of your Basketball-shaped Earth.
When you can NOT find JUST ONE example, please return immediate with an appropriate apology for your confusion over such simple matters.
If, however, you DO find JUST ONE EXAMPLE of an Engineering Calculation (of any type structure, anywhere on Earth) that takes into consideration the "curvature" of your Basketball-shaped Earth, I will glad apologize for the "ignorance" you have repeatedly claimed I have.
I eagerly await your response:
PROVE IT!!!
The following italicised paragraphs are taken from the Wikipedia article on the Humber Bridge
Bridge statistics
The bridge's surface takes the form of a dual carriageway with a lower-level foot and cyclepath on both sides. . There is a permanent 50 mph (80 km/h) speed limit on the full length of the bridge.
Each tower consists of a pair of hollow vertical concrete columns, each 155.5 m (510 ft) tall and tapering from 6 m (20 ft) square at the base to 4.5 × 4.75 m (14.8 × 15.6 ft) at the top. The bridge is designed to tolerate constant motion and bends more than 3 m (10 ft) in winds of 80 mph (129 km/h). The towers, although both vertical, are 34 mm (1.3 inches) farther apart at the top than the bottom due to the curvature of the earth.[8] The total length of the suspension cable is 71,000 km (44,000 miles). The north tower is on the bank, and has foundations down to 8 m (26 ft). The south tower is in the water, and descends to 36 m (118 ft) as a consequence of the shifting sandbanks that make up the estuary.
The following paragraph is from the wikipedia article on the Verrazano-Narrows Bridge.
Because of the height of the towers (693 ft or 211 m) and their distance apart (4,260 ft or 1,298 m), the curvature of the Earth's surface had to be taken into account when designing the bridge—the towers are 1 5⁄8 inches (41.275 mm) farther apart at their tops than at their bases.
This is a link from a website dealing with civil engineering. It concerns compensationg for the curvature of the earth:
http://www.aboutcivil.org/curvature-and-refraction.html (http://www.aboutcivil.org/curvature-and-refraction.html)
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I am here to defend truth.
What do you think of my arguments so far? (remember, I'm a brand-new FEer)
I skipped a couple of pages in this thread. So I don't know if you have gotten the drop on RE figured out yet.
8" over the first mile, much more drop in proceeding miles.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
This chart shows the drop, and how to figure it out.
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I am here to defend truth.
What do you think of my arguments so far? (remember, I'm a brand-new FEer)
I skipped a couple of pages in this thread. So I don't know if you have gotten the drop on RE figured out yet.
8" over the first mile, much more drop in proceeding miles.
http://www.sacred-texts.com/earth/za/za05.htm (http://www.sacred-texts.com/earth/za/za05.htm)
This chart shows the drop, and how to figure it out.
Just a quick note on that chart; it's only good up to a point, and the same applies to the accompanying calculation. If you're going to calculate the total drop over a given distance, there are two ways to do it properly:
For drop perpendicular to the tangent line (as in my diagram earlier in this thread, with drops 'a', 'b', 'c', and 'd'):
drop = radius - (cosine((distance / circumference) * 360) * radius)
For drop perpendicular to the surface at the given distance (as in the diagram linked to by hoppy):
drop = (radius / (cosine((distance / circumference) * 360)) - radius
Distance and circumference must be in the same units, but radius can be any units you like, just remember that the equation you use will give an answer in those units (so if you want an answer in feet, use the radius of the earth in feet: 20,925,524.9). Also, a quick note on the second equation: it will only give a meaningful answer if ((distance / circumference) * 360) < 90.
To anyone about to jump in and rave on about this being "bullsh*t fake maths made up to confuse the general population", I remind you that this is basic trigonometry as taught in high schools, and can be derived from first principles if you put in a little effort (been there, done that).
Enjoy!
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As I have asked before show me "sinking ship on flat and glassy sea with no swell "
Here a set I took a month or two ago.
(http://img15.imageshack.us/img15/747/jt8n.png)
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I did say sinking, not shrinking.
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I did say sinking, not shrinking.
That's what you get when something is moving away from you! :) Compare the width to the height though, and you'll notice the height decreasing more rapidly than the width, and the hull being almost completely obscured by the 3rd frame. Pity about the mirage though... 29silhouette, do you think you'd get an opportunity to take a few more shots at a time when the atmospheric effects aren't so prevalent?
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I did say sinking, not shrinking.
That's what you get when something is moving away from you! :) Compare the width to the height though, and you'll notice the height decreasing more rapidly than the width, and the hull being almost completely obscured by the 3rd frame. Pity about the mirage though... 29silhouette, do you think you'd get an opportunity to take a few more shots at a time when the atmospheric effects aren't so prevalent?
I wonder if they'd mind if someone went and painted horizontal stripes on the pillars of the bridge?
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When I see sinking ships there is swell and I can see the ships large and with clarity with the naked eye like this.
http://upload.wikimedia.org/wikipedia/commons/e/e4/Aground_Tauranga_Pukehina_5_Oct_11_4.jpg (http://upload.wikimedia.org/wikipedia/commons/e/e4/Aground_Tauranga_Pukehina_5_Oct_11_4.jpg)
http://www.rgbstock.com/bigphoto/nDpW942%2Focean (http://www.rgbstock.com/bigphoto/nDpW942%2Focean)
And when there is no swell I see ships like this.
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When I see sinking ships there is swell and I can see the ships large and with clarity with the naked eye like this.
http://upload.wikimedia.org/wikipedia/commons/e/e4/Aground_Tauranga_Pukehina_5_Oct_11_4.jpg (http://upload.wikimedia.org/wikipedia/commons/e/e4/Aground_Tauranga_Pukehina_5_Oct_11_4.jpg)
http://www.rgbstock.com/bigphoto/nDpW942%2Focean (http://www.rgbstock.com/bigphoto/nDpW942%2Focean)
And when there is no swell I see ships like this.
It could just be me, but there appears to be more swell in the 2nd photo...
It would also appear that the horizon is closer than the ships in every case except perhaps the middle ship in the 2nd photo, an impossibility if the earth were flat.
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I did say sinking,
It is 'sinking'. It's disappearing from the bottom up. The water is flat, and not glassy, but it was glassy near shore and I could see the bridge reflected across the water. If it ever is 'glassy' to the horizon, it'll be like trying to look across a mirror anyway.
not shrinking.
It's moving away from me. What effect is supposed to take place then?
29silhouette, do you think you'd get an opportunity to take a few more shots at a time when the atmospheric effects aren't so prevalent?
It's pretty hit or miss with the atmospheric problems. I was looking toward the bridge once, saw a bulge in the water, was wondering what the heck I was looking at, and then noticed the bridge above also had a big upward bump. It was moving and flattening out. Just a random gust of different temperature air.
Hopefully I can catch that tug and barge again, and spend the time to track it further. Had a few other pictures of a ferry, that bridge, and hillside from two different elevations from the same time I took those tug photos. I'll see if thread is still here (might have been deleted with the server mess appears that's been restored) or if imageshack is working enough to let me see my pictures I uploaded.
How on Earth could gravity hold water in a fixed curvature around a basketball shaped object? It just doesn't make sense and is downright incomprehensible.
...verbally abuse knuckleheads who buy into the insane idea that this thing we live on is shaped like a basketball and water is curved.
So you have never looked at a desktop globe, and are unable to visualize water on the surface of said sphere, yet those who can, are the knuckleheads?